Question

If $u_r=a_{r}x+b_{r}y+,c_r=0$ (r = 1 , 2, 3 ) be the three sides of a triangle them the equations of the circumcircle of this triangle is $\left|\begin{array}{ccc}\frac{1}{u_1}& \frac{1}{u_2}& \frac{1}{u_3}\\ a_2{a}_3-b_2{b}_3& a_3{a}_1-b_3{b}_1& a_1{a}_2-b_1{b}_2\\ a_2{b}_3+a_3{b}_2& a_3{b}_1+a_1{b}_3& a_1{b}_2+a_2{b}_1\end{array}\right|$ = 0

Answer 1

It is exactly the same as last.part
$u_2{u}_3+\lambda u_3{u}_1+\mu u_1{u}_2$ = 0. ..(1)
$(a_{2}x+b_{2}y+c_2)(a_{3}x+b_{3}y+c_3)$
+ $\lambda (a_{3}x+b_{3}y+c_3)(a_{1}x+b_{1}y+c_1)$
+ $\mu (a_{1}x+b_{1}y+c_1)(a_{2}x+b_{2}y+c_2)=0$
Apply the condition that it represents a circle
$a_2{a}_3+\lambda a_3{a}_1+\mu a_1{a}_2=b_2{b}_3+\lambda b_3{b}_1+\mu b_1{b}_2$
or $\left(a_2{a}_3{b}_2{b}_3\right)+\lambda \left(\right)+\mu \left(\right)=0$ ..(2)
and $(a_2{b}_3+a_3{b}_2)+\lambda \left(\right)+\mu \left(\right)=0$ ..(3)
Eliminating $\lambda ,\mu$ between (1), (2), (3), we get the result in determinant form. Divide $R_{1}by{u}_1{u}_2{u}_3$ etc.
Another form :- Expanding above the coefficient of $\frac{1}{u_1}$ is
$(a_3{a}_1-b_3{b}_1)(a_1{b}_2+a_2{b}_1)-(a_3{b}_1+a_1{b}_3)(a_1{a}_2-b_1{b}_2)$
There will be eight terms in the above out of which four will cancel and the remaining four are
$a_3{b}_2(a_1^2,+b_1^2)-a_2{b}_3(a_1^2,+b_1^2)$
= $(a_1^2,+b_1^2)(a_3{b}_2-a_2{b}_3)$
= $(a_1^2,+b_1^2)\left|\begin{array}{cc}{a}_2& a_3\\ b_2& b_3\end{array}\right|$etc